Solve y=-(x-16)²: Finding Values Where Function is Negative

Given the function:

$y=-\left(x-16\right)^2$

Determine for which values of X the following is true:

$f(x) < 0$

To solve this problem, we'll analyze the given quadratic function:

• The function is $y = -\left(x - 16\right)^2$, a downward-opening parabola with vertex at $(16, 0)$.
• The quadratic function is in vertex form $y = -\left(x - 16\right)^2$. Here, the value of $y$ is zero when $x = 16$.
• Since $-\left(x - 16\right)^2$ represents a downward-opening parabola, $y$ will be zero only when $x = 16$.
• For $y$ to be less than zero, $x$ must be any real number except 16, as the squared term results in zero exactly when $x = 16$.
• Thus, the inequality $f(x) < 0$ holds for all $x$ except at $x = 16$.

Therefore, the solution to the problem is $x\ne16$.